In Δ A B C ,, K and L are points on BC. AL is the bisector of ∠ K A C. K L × B C = B K × C L. Find ∠ B A L.

Question

In   Δ  A  B  C  ,, K and L are points on BC. AL is the bisector of   ∠  K  A  C.   K  L  ×  B  C  =  B  K  ×  C  L. Find   ∠  B  A  L.

Julien Zuniga · Accepted Answer

Step 1The identity   K  L  ⋅  B  C  =  B  K  ⋅  C  L implies that line AB is the external bisector of triangle CKA at A while AL is the interior bisector at A.Step 2Hence they are orthogonal and thus the angle   ∠    B  A  L  =      90          ∘      .

Hagman7v · Accepted Answer

Step 1By the bisector theorem we have             A      C              A      K        =            C      L              K      L        =            B      C              B      K       hence   μ  =            A      C              C      B        =            A      K              K      B       and K belongs to the Apollonius circle that is the locus of points P such that             P      A              P      B        =  λ. Such a circle goes through the feet of the angle bisectors (the internal and the external) from C. If ABC is an acute-angled triangle, such a circle intersects the BC segment only at C, hence there is no solution.On the other hand, if we allow                     B        A        C            ^       to be greater than       90    ∘  , we have a solution at                     B        A        L            ^        =            90      ∘      :

In triangle ABC, K and L are points on BC. K is closer to B than L. AL is the bisector of angle KAC. KL times BC=BK times CL. Find angle BAL.

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